Figure It Out — Questions & Solutions

Given

Check congruence of the two bent, two-arm figures shown in the textbook.

To Find

Whether the two source angle figures are congruent

Reasoning

A rigid motion preserves both arm lengths and their included angle.

Solution

1. Align one arm and the joint of the two figures.

2. The other arm opens at a different angle in the source drawing, so it does not coincide.

   Equal or similar arm lengths alone are insufficient.

Result

The two displayed figures are not congruent.

Given

Circle the source pairs that appear congruent: teardrops, clouds, starbursts and two-leaf figures.

To Find

Pairs that appear to have the same shape and size

Reasoning

Rotation and reflection are allowed; resizing is not.

Solution

1. The two teardrops can be matched by changing orientation

   they appear congruent.

2. The two-leaf figures appear to match after reflection.

3. The clouds and starbursts have different sizes, so those pairs do not appear congruent.

4. The question asks about appearance.

   For an exact physical check, trace and superimpose the outlines without stretching.

Result

Circle the teardrop pair and the two-leaf pair.

Given

A circle or a rectangle to be copied.

To Find

Sufficient measurements for congruence

Property

A radius fixes a circle; two adjacent side lengths fix a rectangle.

Solution

1. For circles, compare radii.

2. For rectangles, compare the unordered pair of length and width

   rotation may swap their orientation.

3. Area alone is insufficient: 2 × 6 and 3 × 4 have the same area but different dimensions.

Result

Use radius for a circle, and length and width for a rectangle.

Given

State how to check congruence of the three-arm symbol, then assess the two source pairs.

To Find

Required measurements and congruence of both displayed pairs

Reasoning

Measure all arms from the junction and their relative angles.

Solution

1. Record the three arm lengths, the straight-line relation of the two opposite arms, and the angle of the third arm to a specified arm.

2. For each displayed pair, the second symbol is a translated copy of the first: the same endpoints and junction align after shifting.

   Both pairs are congruent.

3. The stored source’s line geometry confirms this equality

   no scale change is required.

Result

Both source pairs are congruent; arm lengths and angles establish the match.

Given

Given △HEN ≅ △BIG, list all other correctly ordered forms of this correspondence.

To Find

Equivalent triangle-order statements

Reasoning

Keep H↔B, E↔I and N↔G in matching positions.

Solution

1. Starting with each vertex and traversing either direction gives six orders in total.

2. Apart from HEN≅BIG, the remaining orders are HNE≅BGI, EHN≅IBG, ENH≅IGB, NHE≅GBI and NEH≅GIB.

3. Each equality can also be read with its two sides reversed

   this does not create a different correspondence.

Result

△HNE≅△BGI; △EHN≅△IBG; △ENH≅△IGB; △NHE≅△GBI; △NEH≅△GIB.

Given

RE = JA, ED = AM and RD = JM.

To Find

The congruence order

Property

SSS.

Solution

1. R is where the 3.5 cm and 6 cm sides meet

   it matches J.

2. E matches A, and D matches M.

Result

△RED ≅ △JAM.

Given

AB = AD, CB = CD and common side AC.

To Prove

∠BAC = ∠CAD and ∠BCA = ∠ACD

Property

SSS then corresponding angles.

Proof

1. AB = AD, BC = DC and AC = AC give △ABC ≅ △ADC.

2. A ↔ A, B ↔ D, C ↔ C.

3. The two angles adjacent to AC at A match

   so do those at C.

Result

AC bisects both specified angles.

Given

DF = DG, FE = GE and common side DE.

To Find

The correctly ordered congruence

Property

SSS with D ↔ D, F ↔ G and E ↔ E.

Solution

1. DF corresponds to DG

   FE corresponds to GE

   DE corresponds to DE.

2. The correct statement is △DFE ≅ △DGE.

3. △DFE ≅ △GED would wrongly pair DF with GE, which is not guaranteed.

Result

The triangles are congruent, but △DFE ≅ △GED is not a justified order.

Given

Two corresponding sides with the included angles B and Z.

To Find

The congruence order

Property

SAS.

Solution

1. B ↔ Z lies between the 7 cm and 5 cm sides.

2. A ↔ X and C ↔ Y.

Result

△ABC ≅ △XZY.

Given

A,O,C and B,O,D are straight lines; AB ∥ CD and AB = CD.

To Find

Corresponding equal parts of the crossing triangles

Property

Alternate interior angles and ASA.

Solution

1. ∠BAO = ∠DCO

   ∠ABO = ∠CDO.

2. AB = CD, so △ABO ≅ △CDO by ASA.

3. The correspondence is A ↔ C, B ↔ D, O ↔ O.

Result

AO = CO, BO = DO and ∠AOB = ∠COD.

Given

The two angle pairs and common side BC.

To Prove

∠BAC = ∠BDC and △ABC ≅ △DBC

Property

Angle sum and ASA.

Proof

1. The third angles are each 180° minus the same two base angles.

2. The known angles have common included side BC.

Result

△ABC ≅ △DBC; the third angles are equal.

Given

B and C are the common base endpoints; the two given angle equalities.

To Find

The correspondence between △ABC and △DCB

Property

Angle addition followed by ASA.

Solution

1. ∠ABC = ∠ABD + ∠DBC

   ∠DCB = ∠DCA + ∠ACB.

   Hence ∠ABC = ∠DCB.

2. ∠ACB = ∠DBC and BC = CB.

3. Thus △ABC ≅ △DCB.

Result

AB = DC, AC = DB and ∠BAC = ∠CDB.

Given

Given △AIR ≅ △FLY, identify corresponding vertices, sides and angles.

To Find

The complete correspondence

Reasoning

The order of a congruence statement fixes corresponding positions.

Solution

1. Vertices: A↔F, I↔L, R↔Y.

2. Sides: AI = FL, IR = LY, RA = YF.

3. Angles: ∠IAR = ∠LFY, ∠AIR = ∠FLY, ∠IRA = ∠LYF.

Result

All three vertex, side and angle pairs follow the stated order.

Given

(a) AB = DE, BC = EF, CA = DF; (b) AB = EF, ∠A = ∠E, AC = ED; (c) AB = DF, ∠B = ∠D = 90°, AC = FE; (d) ∠A = ∠D, ∠B = ∠E, AC = DF; (e) AB = DF, ∠B = ∠F, AC = DE

To Find

Congruence criterion and valid order in cases (a)–(e)

Reasoning

Use only a sufficient criterion; SSA does not guarantee congruence.

Solution

1. (a) SSS: The endpoints of AB and AC give A ↔ D.

   Then B ↔ E and C ↔ F. △ABC ≅ △DEF.

2. (b) SAS: The included angles A and E lie between the given side pairs

   A ↔ E, B ↔ F, C ↔ D. △ABC ≅ △EFD.

3. (c) RHS: AC and FE are hypotenuses

   the right-angle vertices B and D correspond.

   AB ↔ FD gives A ↔ F and C ↔ E. △ABC ≅ △FDE.

4. (d) AAS: The third angles C and F agree by the angle sum, so AAS applies. △ABC ≅ △DEF.

5. (e) SSA — insufficient in general: The angle is not included between AB and AC, or between DF and DE.

   The two-intersection construction is a counterexample.

   No congruence guaranteed.

Result

(a) △ABC≅△DEF; (b) △ABC≅△EFD; (c) △ABC≅△FDE; (d) △ABC≅△DEF; (e) no guaranteed congruence.

Given

A,O,D are collinear and B,O,C are collinear; O lies between the endpoints; OA = OD and OB = OC.

To Prove

AB ∥ CD

Property

SAS followed by the converse of the alternate-interior-angle property.

Proof

1. ∠AOB = ∠DOC are vertically opposite angles.

2. OA = OD and OB = OC are given, so △AOB ≅ △DOC by SAS.

3. Corresponding angles ∠BAO and ∠CDO are equal.

4. These are alternate interior angles for AB and CD with transversal AD.

Result

AB ∥ CD. Also AB = DC by corresponding sides.

Given

Square ABCD split along AC.

To Find

The valid correspondences and an example with six matches

Property

SSS; permutations must preserve every side length.

Solution

1. AB = AD, BC = DC and AC is common: △ABC ≅ △ADC.

2. AB = CD, BC = DA and AC = CA: △ABC ≅ △CDA.

3. The right-angle vertex B must match D, so these square halves have two matches, not six.

4. Take two equilateral triangles ABC and XYZ of the same side length.

   Every vertex permutation preserves all side lengths.

5. With ABC fixed, valid orders are XYZ, XZY, YXZ, YZX, ZXY and ZYX.

6. Another pair with two matches is any two congruent non-equilateral isosceles triangles.

   Match their apex vertices and either pair their base endpoints in order or interchange those endpoints.

Result

Two matches for the square halves; six for a pair of congruent equilateral triangles.

Given

A is the centre; B and C lie on the circle; ∠BAC = 120°.

To Find

The measures of ∠ABC and ∠ACB

Property

Equal radii, isosceles base angles and the triangle angle sum.

Solution

1. AB = AC because both are radii.

2. Therefore ∠ABC = ∠ACB.

3. Their sum is 180° − 120° = 60°, so each is 30°.

Result

∠ABC = ∠ACB = 30°. This uses the 120° interior central angle, not its reflex angle.

Given

Find the missing angles in the source angle network, respecting the groups of equal-length marks.

To Find

Valid local angle deductions and consistency of the full figure

Reasoning

Isosceles base angles, straight angles and triangle angle sums; distinguish local deductions from a globally exact construction.

Solution

1. The source data are not jointly exact: equal single-marked lengths give outer rectangle sides in ratio 3:2, whose diagonal makes arctan($\frac{2}{3}$)≈33.69°, not exactly the printed 34°.

2. Accordingly, no globally consistent exact assignment satisfies all the source marks.

   The following values are conditional local deductions, not a claim that the full network exists exactly.

3. M,N,O,P label the source’s unnamed interior points as in the accompanying diagram.

4. △CUR, angles in that vertex order: 90°, 45°, 45°.

   CU = CR

   angle C is right.

5. △AUN, angles in that vertex order: 56°, 68°, 56°.

   AU = UN

   use the angle sum.

6. △UMN, angles in that vertex order: 34°, 73°, 73°.

   UM = UN

   the two base angles share 146°.

7. △RVO, angles in that vertex order: 56°, 68°, 56°.

   RV = VO

   the two base angles share 112°.

8. △VOD, angles in that vertex order: 112°, 34°, 34°.

   Straight line RVD gives 112° at V

   VO = VD.

9. △DOF, angles in that vertex order: 56°, 26°, 98°.

   Corner D is 90°

   subtract 34°, then use the triangle sum.

10. △PFB, angles in that vertex order: 60°, 60°, 60°.

    PF = FB = BP are double-marked.

11. △KLP, angles in that vertex order: 30°, 90°, 60°.

    KL = LB and common perpendicular LP give △KLP ≅ △BLP by SAS

    K,P,F are collinear.

12. △PLB, angles in that vertex order: 60°, 90°, 30°.

    The equilateral angle at B leaves 30° in the right corner.

13. △NOP, angles in that vertex order: 34°, 90°, 56°.

    Use the two given angles and the angle sum.

14. △ANK, angles in that vertex order: 34°, 44°, 102°.

    Two printed angles give the third.

15. △NKP, angles in that vertex order: 102°, 48°, 30°.

    AN and NO are opposite rays, so ∠KNO = 136°

    subtract ∠PNO = 34°.

16. △PFO, angles in that vertex order: 94°, 22°, 64°. ∠OFB = 82° and ∠PFB = 60°

    then use ∠NPO = 56° and ∠NPK = 30° with K,P,F collinear.

17. △MNO, angles in that vertex order: 83°, 51°, 46°.

    At N: 180° − 56° − 73° = 51°

    at O the printed angle is 46°.

18. △MRO, angles in that vertex order: 102°, 34°, 44°.

    Two printed angles give the third.

19. △URM, angles in that vertex order: 33°, 45°, 102°.

    At R on CRV: 180° − 45° − 34° − 56° = 45°

    at U on CUA: 180° − 45° − 34° − 68° = 33°.

Result

Local deductions are listed with their assumptions; a corrected source figure is required for a globally exact solution.